What are the solutions to the equation 7(4x-3)^2-15=-68 ?

The solutions to the equation 7(4x-3)^2-15=-68 are x= 3/4+i(sqrt(371))/(28),x= 3/4-i(sqrt(371))/(28)

Find the zeros of 7(4x-3)^2-15=-68

The zeros of 7(4x-3)^2-15=-68 are x= 3/4+i(sqrt(371))/(28),x= 3/4-i(sqrt(371))/(28)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

7 (4 x - 3)^{2} - 15 = - 68

x = \frac{3}{4} + i \frac{371}{28}, x = \frac{3}{4} - i \frac{371}{28}

Solution steps

7 (4 x - 3)^{2} - 15 = - 68

Move

15

to the right side

7 (4 x - 3)^{2} = - 53

Divide both sides by

7

(4 x - 3)^{2} = - \frac{53}{7}

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

4 x - 3 = - \frac{53}{7} : x = \frac{3}{4} + i \frac{371}{28}

Solve

4 x - 3 = - - \frac{53}{7} : x = \frac{3}{4} - i \frac{371}{28}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{371}{28}, x = \frac{3}{4} - i \frac{371}{28}

- 3 x^{2} + 174 x - 200 = 0

12 x^{2} - 24 x - 36 = 0

x^{2} + 6 x - 10

x^{2} - 2 (x - 3) = 6 + 2 x

5 x^{2} - 5 x - 150 = 0

What are the solutions to the equation 7(4x-3)^2-15=-68 ?
The solutions to the equation 7(4x-3)^2-15=-68 are x= 3/4+i(sqrt(371))/(28),x= 3/4-i(sqrt(371))/(28)
Find the zeros of 7(4x-3)^2-15=-68
The zeros of 7(4x-3)^2-15=-68 are x= 3/4+i(sqrt(371))/(28),x= 3/4-i(sqrt(371))/(28)